Probability: complete GMAT guide

The basic definition

For an event $A$ in a sample space of equally likely outcomes,

P(A)=\frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}

A probability is always between 0 and 1. An impossible event has probability 0, and a certain event has probability 1. The sum of the probabilities of all possible outcomes is 1.

Example. A bag contains 4 red, 3 blue, and 5 green marbles. What is the probability of drawing a blue marble?

Total marbles: $4+3+5=12$ . Blue marbles: 3. Probability: $3/12=1/4$ .

The complement rule

The probability that an event does not occur is

P(\text{not } A)=1-P(A)

The complement is often easier to count. If a question asks for the probability of "at least one" success, calculate the probability of no successes and subtract from 1.

Example. A fair coin is flipped 4 times. What is the probability of getting at least one head?

The complement is getting no heads, which means four tails. Probability of four tails: $(1/2)^4=1/16$ . Probability of at least one head: $1-1/16=15/16$ .

Independent events

Two events are independent if the occurrence of one does not change the probability of the other. For independent events,

P(A \text{ and } B)=P(A) \times P(B)

Coin flips are independent. Drawing cards without replacement is not independent, because the first draw changes the composition of the deck.

Example. A fair die is rolled twice. What is the probability of rolling a 6 both times?

$(1/6)(1/6)=1/36$ .

Dependent events

For dependent events, multiply the probability of the first event by the conditional probability of the second event given the first:

P(A \text{ and } B)=P(A) \times P(B|A)

If a bag has 5 red and 3 blue marbles, the probability of drawing red and then blue without replacement is $(5/8)(3/7)=15/56$ . The denominator drops from 8 to 7 because one marble is gone.

Example. Two cards are drawn from a standard deck without replacement. What is the probability that both are aces?

First ace: $4/52$ . Second ace given the first: $3/51$ . Probability: $(4/52)(3/51)=1/221$ .

Mutually exclusive events

Two events are mutually exclusive if they cannot occur at the same time. For mutually exclusive events,

P(A \text{ or } B)=P(A)+P(B)

For events that can overlap, subtract the overlap:

P(A \text{ or } B)=P(A)+P(B)-P(A \text{ and } B)

Drawing a heart or a king from a deck is not mutually exclusive, because the king of hearts belongs to both. The probability is $13/52+4/52-1/52=16/52=4/13$ .

Conditional probability

Conditional probability asks for the probability of one event given that another has occurred:

P(A|B)=\frac{P(A \text{ and } B)}{P(B)}

In a table of 40 students, suppose 24 study math, 18 study physics, and 10 study both. The probability that a student studies math given that the student studies physics is $10/18=5/9$ . The denominator is the group with the condition, not the entire table.

Example. In a class, 60% of students play a sport, 30% play an instrument, and 15% do both. What is the probability that a student plays an instrument given that the student plays a sport?

$P(\text{instrument}|\text{sport})=0.15/0.60=0.25=25\%$ .

Probability with counting

When outcomes are equally likely and involve selections, count with combinations. The probability of choosing 2 women from a group of 4 women and 3 men is $\binom{4}{2}/\binom{7}{2}=6/21=2/7$ . The numerator and denominator must count the same type of outcome.

If the question involves arranging objects, count arrangements in both the numerator and denominator. For example, the probability that a random arrangement of the letters in BANANA starts with B requires counting the total distinct arrangements and the arrangements that begin with B.

Expected value

The expected value of a random variable is the sum of each outcome multiplied by its probability. If a game pays $10 with probability 0.2, $5 with probability 0.3, and $0 with probability 0.5, the expected payout is $10(0.2)+5(0.3)+0(0.5)=3.5$ dollars.

Expected value does not have to be a possible outcome. It is a long-run average, not a prediction of a single trial.

Probability trees

A probability tree organizes multi-step events. Each branch represents a possible outcome, and the probability of a path is the product of the probabilities along it. If a bag contains 3 red and 2 blue marbles, the probability of drawing red then blue without replacement is the product of the first branch $3/5$ and the second branch $2/4$ , which gives $6/20=3/10$ .

Add the probabilities of all paths that satisfy the event. For example, the probability of drawing one red and one blue in either order is the sum of two paths: red then blue, and blue then red. Each path has the same probability, so the total is $2 \times 3/10=3/5$ .

Example. A coin is biased so that heads occurs with probability 0.6. The coin is flipped twice. What is the probability of exactly one head?

There are two paths: heads then tails, with probability $0.6 \times 0.4=0.24$ , and tails then heads, with probability $0.4 \times 0.6=0.24$ . Total: $0.48$ .

Sampling with and without replacement

With replacement, the composition of the pool stays the same, so the probabilities do not change from draw to draw. Without replacement, each draw changes the pool. Read the question for the phrase "without replacement" or "one after another." If the item is put back, use the original denominator each time.

Situation	Denominator
With replacement	stays the same
Without replacement	decreases by one each draw
Drawing simultaneously	treat as without replacement

Drawing two items at the same time is the same as drawing them one after another without replacement. The order does not matter for the probability, but the pool shrinks after the first item is removed.

If an event has probability $p$ , the event occurs at least once in $n$ independent trials with probability $1-(1-p)^n$ . This formula is the complement of the event occurring zero times. It is useful for questions about repeated attempts.

A repeatable approach

Identify the total set of equally likely outcomes.
Decide whether to count directly or use the complement.
Determine whether events are independent, dependent, or mutually exclusive.
For "and," multiply probabilities when appropriate.
For "or," add probabilities and subtract any overlap.
Check that the probability is between 0 and 1.

Common traps

Adding probabilities for events that are not mutually exclusive.
Multiplying probabilities for events that are not independent.
Forgetting to change the denominator when drawing without replacement.
Using the total group instead of the conditional group.
Assuming outcomes are equally likely when they are not.
Counting ordered outcomes in the numerator and unordered outcomes in the denominator.

Quick reference

Rule	Formula
Basic probability	$\text{favorable}/\text{total}$
Complement	$1-P(A)$
Independent "and"	$P(A)P(B)$
Dependent "and"	$P(A)P(B\|A)$
Mutually exclusive "or"	$P(A)+P(B)$
General "or"	$P(A)+P(B)-P(A \text{ and } B)$
Conditional probability	$P(A \text{ and } B)/P(B)$

Probability questions are counting questions with a ratio at the end. Define the total set first, then count the favorable set the same way. If the events overlap, subtract the overlap. If the question says "at least one," try the complement.

Probability: complete GMAT guide

The basic definition

The complement rule

Independent events

Dependent events

Mutually exclusive events

Conditional probability

Probability with counting

Expected value

Probability trees

Sampling with and without replacement

A repeatable approach

Common traps

Quick reference

Practice Problem Solving: Probability

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